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Supplementary Material to Working Paper 08-07
Sectoral vs. Aggregate Shocks: A Structural Factor Analysis of
Industrial Production
Andrew T. Foerster
Department of Economics, Duke University
Pierre-Daniel G. Sarte
Research Department, Federal Reserve Bank of Richmond
Mark W. Watson
Department of Economics, Princeton University
1
1
The Model
max Et
∞
X
βt
t=0
subject to
Yjt = Cjt +
N
X
j=1
N
X
i=1
and
Yjt =
Ã
1−σ
−1
Cjt
− ψLjt
1−σ
!
Mjit + Kjt+1 − (1 − δ)Kjt
α
Ajt Kjtj
N
Y
γ
i=1
The first-order necessary conditions are:
S
1−αj − N
i=1 γ ij
Mijtij Ljt
.
−σ
= λjt ,
Cjt : Cjt
Ã
!
N
X
Yjt
Ljt : ψ = λjt
1 − αj −
γ ij .
Ljt
i=1
Combining these two equations gives
−σ Yjt
ψ = Cjt
Ljt
Ã
1 − αj −
Mijt : λit = λjt γ ij
or
−σ
γ ij
Cit−σ = Cjt
Kjt+1 :
2
−σ
Cjt
N
X
γ ij
i=1
!
.
Yjt
,
Mijt
Yjt
.
Mijt
∙
µ
¶¸
Yjt+1
−σ
= βEt Cjt+1 αj
+1−δ
Kjt+1
Dynamics of the System
The dynamics are described by a set of 4N + N 2 equations in 4N + N 2 unknowns. When
N = 117, this amounts to 14157 equations, but preliminary algebraic manipulations help
keep the system tractable.
Ã
!
N
X
−σ
ψLjt = Cjt
1 − αj −
γ ij Yjt
i=1
−σ
Cit−σ = Cjt
γ ij
2
Yjt
Mijt
−σ
Cjt
∙
µ
¶¸
Yjt+1
−σ
= βEt Cjt+1 αj
+1−δ
Kjt+1
Yjt = Cjt +
N
X
i=1
and
Yjt =
Mjit + Kjt+1 − (1 − δ)Kjt
α
Ajt Kjtj
N
Y
i=1
3
S
γ ij 1−αj − N
i=1 γ ij
Mijt Ljt
.
Log-linearized Equations
The “hat” notation stands for percent deviation from steady state.
bjt + Ybjt
bjt = −σ C
L
bit = −σ C
bjt + Ybjt − M
cijt
−σ C
eYbjt+1 − β
eK
bjt = −σEt C
b jt+1
bjt+1 + β
−σ C
e = 1 − β + βδ.
where β
bjt + SK K
b jt+1 − (1 − δ)SK K
b jt +
Ybjt = SCj C
j
j
bjt + αj K
b jt +
Ybjt = A
N
X
i=1
Ã
cijt + 1 − αj −
γ ij M
N
X
i=1
cjit
SMji M
N
X
γ ij
i=1
!
bjt .
L
b1t , ..., C
bNt ], etc... and mt = [M
c11t , ..., M
c1Nt , M
c21t , ..., M
cNN t ]. The log-linearized
Let ct = [C
equations can be written in matrix form as follows:
lt = −σct + yt ,
(1)
mt = My yt + Mc ct
(2)
where
My = 1N×1 ⊗ I and Mc = σ(I ⊗ 1N×1 ) − σ(1N ×1 ⊗ I),
e t yt+1 − βk
e t+1
−σct = −σEt ct+1 + βE
(3)
yt = Sc ct + Sm mt + Sk kt+1 − Sk (1 − δ)kt ,
(4)
where
⎡
⎢
Sc = ⎣
⎡
⎤
SC1
...
SCN
⎢
⎥
⎦ , etc... and Sm = ⎣
3
SM11 SM12 ...
0
0
0
0
... SMN,N −1 SMN N
⎤
⎥
⎦,
e t + Φlt ,
yt = at + αd kt + Γm
where
eN ×N 2
Γ
⎡
γ 11 0 ... γ 21
⎢
⎢ 0 γ 12 ... 0
⎢
=⎢
γ 13
⎢
⎢
...
⎣
and
γ 1N
⎡ P
⎢
⎢
Φ = I − αd − ⎢
⎢
⎣
4
0
γ 22
i γ i1
(5)
... γ N1 0
...
...
γ N2
γ 23
γ N3
...
...
0
... γ 2N
γ NN
P
i
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
γ i2
...
P
i
System Reduction
γ iN
⎥
⎥
⎥
⎥
⎦
Use equation (1), (2) and (5) to obtain
e
e y − Φ]yt = at + αd kt + [ΓM
− Φσ]ct
[I − ΓM
| c{z
}
{z
}
|
Ωyc
αd
or, equivalently
−1
yt = α−1
d at + kt + αd Ωyc ct .
Note that Ωyc = αd (I − (I − Γ0 )α−1
d ) when σ = 1. Substituting this equation in equation (3)
gives
e −1 at+1 + kt+1 + α−1 Ωyc ct+1 ] − βk
e t+1
−σct = −σct+1 + β[α
d
d
or
e −1 Ωyc ]ct+1 + βα
e −1 at+1 .
−σct = [−σI + βα
d
d
Use the resource constraint (4) to obtain
yt = Sc ct + Sm [My yt + Mc ct ] + Sk kt+1 − Sk (1 − δ)kt
or
(I − Sm My )yt = (Sc + Sm Mc )ct + Sk kt+1 − Sk (1 − δ)kt
which gives
−1
(I − Sm My )[α−1
d at + kt + αd Ωyc ct ] = (Sc + Sm Mc )ct + Sk kt+1 − Sk (1 − δ)kt
4
(7)
or finally
Sk kt+1 = [(I − Sm My )α−1
d Ωyc − (Sc + Sm Mc )]ct + [Sk (1 − δ) + (I − Sm My )]kt
(8)
+(I − Sm My )α−1
d at
We can write equations (7) and (8) as:
# "
#
"
e −1 Ωyc 0
c
−σI + βα
t+1
d
Et
0
Sk
kt+1
#
#"
"
ct
−σI
0
+
=
kt
(I − Sm My )α−1
d Ωyc − (Sc + Sm Mc ) Sk (1 − δ) + (I − Sm My )
"
#
#
"
e −1
0
−βα
d
at +
Et (at+1 )
−1
(I − Sm My )αd
0
(9)
At this stage, the dynamics of the system can be solved using standard linear rational
expectations toolkits such as Blanchard and Kahn (1980), King, Plosser, Rebelo (1988), and
Klein (2000). The results presented in the text are based on King and Watson (2002). To
use these methods, however, one must first obtain the steady state of the system. In this
model, this can be achieved analytically.
5
Finding the Steady State Analytically
The steady state solution only requires inverting N ×N matrices. The steady state equations
for labor, materials, and capital are respectively
ψLj = λj (1 − αj −
M1j =
N
X
γ ij )Yj
i=1
λj
γ Yj
λi ij
1
Kj = αj [ − 1 + δ]−1 Yj .
β
|
{z
}
φKj
Now take the logs of these equations to obtain (small letters denote logs)
lj = − ln ψ + ln λj + ln(1 − αj −
N
X
γ ij ) + yj ,
i=1
mij = ln λj − ln λi + ln γ ij + yj
5
kj = ln φKj + yj .
(12)
The log steady state equations can be written in matrix form to summarize the entire
system. Let l = [l1 , ..., lN ], etc... and m = [m11 , m12 , ...m1N , m21 , ...mNN ]. Then, we have
that
e + ln λ + y,
l = − ln ψ + ln Φ
where
⎡
⎤
P
ln(1 − α1 − i γ i1 )
⎥
e=⎢
ln Φ
...
⎣
⎦.
P
ln(1 − αN − i γ iN )
Similarly, the equation for steady state materials can be expressed as,
m = Mλ ln λ + My y + vec(ln Γ0 ),
where
Mλ = 1N×1 ⊗ I − I ⊗ 1N×1 , and My = 1N×1 ⊗ I.
Finally, we have that
k = ln φK + y,
where
⎡
⎤
ln φK1
⎢
⎥
ln φK = ⎣ ... ⎦
ln φKN
The log of production in all sectors can be expressed as
e + Φl,
y = a + αd k + Γm
e is defined as above. Making the appropriate substitutions yields
where Γ
e λ ln λ + My y + vec(ln Γ0 )] + Φ(− ln ψ + ln Φ
e + ln λ + y)
y = a + αd (ln φK + y) + Γ[M
or, equivalently,
e
e
e − Φ ln ψ + (ΓM
e λ + Φ) ln λ.
[I − αd − ΓM
− Φ]y = a + αd ln φK + Γvec(ln
Γ0 ) + Φ ln Φ
{z y
}
|
=0
It follows that we can solve for (shadow) prices in the steady state in closed form,
e
e − Φ ln ψ],
e λ + Φ)−1 [a + αd ln φK + Γvec(ln
Γ0 ) + Φ ln Φ
ln λ = −(ΓM
and λ = eln λ .
6
To solve for the vector Y , write the resource constraints as
1
λ− σ + δφdK Y + Mr Y = Y
where
φdK
⎡
⎢
⎢
=⎢
⎢
⎣
⎤
φK1
...
φKN −1
ΦKN
⎡
⎥
⎢
⎥
⎢
⎥ , and Mr = ⎢
⎥
⎢
⎦
⎣
γ 11
γ 12 λλ21
γ 21 λλ12
γ 22
...
λ1
γ N 1 λN
γ N2 λλN2
... γ 1N λλN1
... γ 2N λλN2
...
... γ NN
⎤
⎥
⎥
⎥,
⎥
⎦
and φdK is a diagonal matrix with φK on its diagonal. The solution for Y is then given by
1
Y = [I − δφdK − Mr ]−1 λ− σ .
Solving for the remaining variables in the steady state is then straightforward.
6
Output from King and Watson (2002) programs
The policy functions take the form (with 2 sectors as an example):
⎤⎡
⎡
⎤ ⎡
⎤
k1t
...
c1t
⎥⎢
⎢
⎥ ⎢
⎥
⎥ ⎢ k2t ⎥
⎢ c2t ⎥ ⎢
⎥⎢
⎢
⎥=⎢
⎥
⎢ k ⎥ ⎢ 1 0 0 0 ⎥⎢ δ ⎥,
⎦ ⎣ 1t ⎦
⎣ 1t ⎦ ⎣
0 1 0 0
k2t
δ 2t
"
# "
a1t
...
=
a2t
| {z }
1 0
0 1
xt
and
⎡
⎢
⎢
⎢
⎢
⎣
k1t+1
k2t+1
δ 1t+1
δ 2t+1
k1t
k2t
δ 1t
δ 2t
#⎢
⎢
⎢
⎢
⎣
⎡
⎤
⎡
#⎢
⎥ "
⎢
⎥
M
M
k
a
⎢
⎥=
⎢
⎥
0
I
⎣
⎦
k1t
k2t
δ 1t
δ 2t
⎤
⎤
⎥
⎥
⎥,
⎥
⎦
⎥
⎥
⎥ + Hεt .
⎥
⎦
More generally, we can write these equations as
# "
#
"
#"
Πck Πca
ct
kt
=
,
kt
I
0
δt
| {z }
st
7
xt =
and,
7
"
kt+1
δ t+1
#
=
"
h
Q
Mk Ma
0
I
i
st
#"
kt
δt
#
+ Hεt .
Obtaining the Filtering Matrices
Since we assume that the logarithm of sectoral productivity follows a random walk, Q = I
in the procedure governing the driving process (i.e. drp.gss) of King and Watson (2002).
Then, we have that
kt+1 = Mk kt + Ma at
while
ct = Πck kt + Πca at .
Recall that
−1
yt = α−1
d at + kt + αd Ωyc ct .
Therefore,
−1
yt = α−1
d at + kt + αd Ωyc [Πck kt + Πca at ]
[I + Ωyc Πca ]at + [I + α−1
= α−1
d Ωyc Πck ]kt
{z
}
{z
}
|d
|
Πa
so that
Πk
−1
kt = Π−1
k yt − Πk Πa at .
Using these equations, we have that
yt+1 = Πk kt+1 + Πa at+1
= Πk (Mk kt + Ma at ) + Πa at+1
−1
= Πk Mk (Π−1
k yt − Πk Πa at ) + Πk Ma at + Πa at+1
or
yt+1 = Πk Mk Π−1
y + Π (M − M Π−1 Π )a + Πa at+1 .
| {z k } t | k a {z k k a} t
%
Ξ
Under the assumptions made in the paper regarding the process for at , it follows that
∆yt+1 = %∆yt + Ξεt + Πa εt+1 ,
8
so that the filtering is carried out according to
−1
−1
εt+1 = Π−1
a ∆yt+1 − Πa %∆yt − Πa Ξεt .
where ε0 is set to zero.1
Let
η t+1 = Ξεt + Πa εt+1 ,
Then, if var(εt ) = I,
Σηη = ΞΞ0 + Πa Π0a .
References
[1] Blanchard, O. and C. Kahn (1980), “The solution of linear difference models under
rational expectations, Econometrica, 48:1305-1311.
[2] King, R. G., Plosser, C. I, and S. T. Rebelo (1988), “Production, growth, and business
cycles, technical appendix,” manuscript, University of Rochester.
[3] King, R. G., and M. W. Watson (2002), “System reduction and solution algorithms for
singular linear difference systems under rational expectations,” 20:57-86.
[4] Klein, P. (2000), “Using the generalized Schur form to solve a multivariate linear rational
expectations model,” Journal of Economic Dynamics and Control, 24:1405-1423.
1
For the various calibrations presented in the text, the eigenvalues of Π−1
a Ξ have modulus less than one.
9
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